Difference between revisions of "Aufgaben:Problem 8"

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(Solution (b): symmetry should be enough for C_2)
 
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==Task==
  
 +
Let \(R > 0\). On \(\mathbb{R}^2\), consider the metric
 +
$$ g(u,v) = \frac{4R^4}{(R^2+u^2+v^2)^2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\ (u,v) \in \mathbb{R}^2$$
  
''Concerning part a)''
+
a) Write down the Hamiltonian equations describing the geodesic curves.
  
It is kind of unclear to me how \( \frac{d}{dx} ln(x) \vert_{x=1} = -x\dfrac{(\ln(1)-\ln(1-\frac{x}{k}))}{\frac{x}{k}} \). Wouldn't it be much easier to just:
+
b) Let
 +
$$ C_1 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})$$
 +
$$ C_2 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{v} +(R^2+u^2-v^2)\dot{u})$$
 +
$$ C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v})$$
 +
Show that \(C_1, C_2, C_3\) are conserved and that
  
$$ \lim_{k \rightarrow \infty} \ln((1-\frac{x}{k})^{k}) = \lim_{k \rightarrow \infty}  k\ln(1-\frac{x}{k}) = \lim_{k \rightarrow \infty}  \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} $$
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$$ -C_1 u + C_2 v +C_3(u^2+v^2-R^2) = 0 \forall (u,v)\in \mathbb{R}^2$$
 +
c) Show that the energy h satisfies \(h \geq 0\) and discuss the case \(h = 0\).
  
and then use Bernoulli-L'Hopital to get:
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d) Describe the geodesics when \(h > 0\) depending on the values of \(C_1, C_2, C_3\).
  
$$ \lim_{k \rightarrow \infty}  \frac{\ln(1-\frac{x}{k})}{\frac{1}{k}} = \lim_{k \rightarrow \infty}  \frac{\frac{x}{k^2}}{(\frac{x}{k} - 1) \frac{1}{k^2}} = -x $$
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==Solution==
  
Also I think you might mean this one:
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===Solution (a)===
  
$$ ((1-\dfrac{x}{k})^{k}\cdot ... \cdot (1 - x) \cdot 1)^{\frac{1}{k+1}} \leq^{AM-GM} \dfrac{1}{k+1}((1-\dfrac{x}{k})^k + ... + (1 - x) +1) \leq \dfrac{1}{k+1}((1-\dfrac{x}{k}) + ... + (1 - x) +1) $$  
+
$$ g_{ij}(u,v) = \frac{4R^4}{(R^2+u^2+v^2)^2} \delta_{ij}$$
  
where we took away the exponents since \( n \in \mathbb{N}, x \in \left( 0, n \right) : (1 - \frac{x}{n}) < 1 \Rightarrow (1 - \frac{x}{n})^n \leq (1 - \frac{x}{n}) \). (Here I actually see a problem, since this would generally hold for \( x \in \left( 0, 1 \right) \) and not on [0, n]) So:
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because \( \big(g^{ij}(u,v)\big) =  \big(g_{ij}(u,v)\big)^{-1}\)
 +
$$ \Rightarrow g^{ij}(u,v)  = \frac{(R^2+u^2+v^2)^2}{4R^4} \delta^{ij}$$
  
$$ ((1-\dfrac{x}{k})^{k}\cdot ... \cdot (1 - x) \cdot 1)^{\frac{1}{k+1}} \leq \frac{1}{k+1}(k(1-\dfrac{x}{k})+1) = \dfrac{1}{k+1}(k + 1 - x) = (1 - \dfrac{x}{k+1}) $$
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let \( x = (u,v)\) and \( \xi = (\xi_1, \xi_2)\), the hamiltonian is defined as
 +
$$h(x, \xi) = \frac{1}{2} g^{ij}(x)\xi_i \xi_j $$
  
Otherwise wouldn't you have to use:
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$$ \Rightarrow h(x, \xi) = \frac{(R^2+u^2+v^2)^2}{8R^4} (\xi_1^2 + \xi_2^2)$$
  
$$ ((1-\dfrac{x}{k})^{k}\cdot1)^{\frac{1}{2}} \leq \dfrac{1}{2}(k(1-\dfrac{x}{k})^k+1) $$
+
now we can compute the Hamiltonian equations:
  
to use the GM-AM?
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$$ \dot{u} = \frac{\partial h}{\partial \xi_1} = \frac{(R^2+u^2+v^2)^2}{4R^4} \xi_1$$
 +
$$ \dot{v} = \frac{\partial h}{\partial \xi_2} = \frac{(R^2+u^2+v^2)^2}{4R^4} \xi_2$$
  
Greetings
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$$ \dot{\xi_1} = -\frac{\partial h}{\partial u} = -\frac{(R^2+u^2+v^2)}{2R^4} (\xi_1^2 + \xi_2^2) u$$
A.
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 +
$$ \dot{\xi_2} = -\frac{\partial h}{\partial v} = -\frac{(R^2+u^2+v^2)}{2R^4} (\xi_1^2 + \xi_2^2) v$$
 +
 
 +
we can also observe that:
 +
 
 +
$$ \dot{\xi_2}u = \dot{\xi_1}v$$
 +
and
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$$ \dot{u}\xi_2 = \dot{v}\xi_1$$
 +
 
 +
 
 +
 
 +
===Solution (b)===
 +
 
 +
$$ C_1 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{u} +(R^2-u^2+v^2)\dot{v}) $$
 +
using (a)
 +
$$ = 2uv\xi_1 +(R^2 -u^2 +v^2)\xi_2  = 2uv\xi_1 +R^2\xi_2 -u^2\xi_2 +v^2\xi_2$$
 +
 
 +
if \(C_1\) is conserved, its time derivative should be zero
 +
 
 +
$$\frac{dC_1}{dt} = 2\dot{u}v\xi_1 + 2u\dot{v}\xi_1 + 2uv\dot{\xi_1}+R^2\dot{\xi_2} - 2u\dot{u}\xi_2 - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$
 +
using \(\dot{v}\xi_1= \dot{u}\xi_2\) on the second term, the second and fifth term cancel:
 +
$$ = 2\dot{u}v\xi_1 +2uv\dot{\xi_1}+R^2\dot{\xi_2} - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$
 +
using \(v\dot{\xi_1} = u\dot{\xi_2}\) on the second term:
 +
$$ = 2\dot{u}v\xi_1 +2u^2\dot{\xi_2}+R^2\dot{\xi_2} - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$
 +
$$ = \dot{\xi_2}(R^2 + u^2+  v^2) + 2\dot{u}v\xi_1 + 2v\dot{v}\xi_2 $$
 +
using the Hamiltonian equations:
 +
$$ = -\frac{(R^2+u^2+v^2)^2}{2R^4}(\xi_1^2 + \xi_2^2) v + 2v\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_1^2 + 2v\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_2^2 =0$$
 +
 
 +
As \(C_1, C_2\) and the hamilton equation are symmetric under permutation of \(u\) and \(v\), \(C_2\) is also conserved.
 +
 
 +
and for \( C_3 \)
 +
 
 +
$$ C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v}) = (v\xi_1 - u\xi_2)$$
 +
$$ \frac{dC_3}{dt} = \dot{v}\xi_1 + v\dot{\xi_1} - \dot{u}\xi_2 - u\dot{\xi_2} = \dot{u}\xi_2 + v\dot{\xi_1} - \dot{u}\xi_2 - v\dot{\xi_1} = 0$$
 +
 
 +
Last we want to prove the identity:
 +
 
 +
$$ -C_1 u + C_2 v +C_3(u^2+v^2-R^2) = \frac{4R^4}{(R^2+u^2+v^2)^2}\Big(-(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})u $$
 +
$$ + (2uv\dot{v} +(R^2+u^2-v^2)\dot{u})v + (v\dot{u} - u\dot{v})(u^2+v^2-R^2)\Big)$$
 +
$$ = \frac{4R^4}{(R^2+u^2+v^2)^2}\Big( -2uv\dot{u}u -R^2\dot{v}u+u^2\dot{v}u-v^2\dot{v}u  + 2uv\dot{v}v +R^2\dot{u}v+u^2\dot{u}v-v^2\dot{u}v$$
 +
$$+ v\dot{u}u^2 + v\dot{u}v^2 - v\dot{u}R^2 - u\dot{v}u^2 - u\dot{v}v^2 + u\dot{v}R^2\Big) = 0$$
 +
as all the terms in the second bracket cancel.
 +
 
 +
===Solution (c)===
 +
 
 +
$$ h(x, \xi) = \frac{(R^2+u^2+v^2)^2}{8R^4} (\xi_1^2 + \xi_2^2)$$
 +
\(h \geq 0\) simply because all the terms are squared. For h = 0 the last bracket has to be zero because \(R > 0\) \( \Rightarrow \xi_1^2 = - \xi_2^2 \Rightarrow \xi_1 = \xi_2 = 0 \) from the Hamiltonian equation follows that u = const. , v = const. which corresponds to the case of a constant geodesic.
 +
 
 +
 
 +
===Solution (d)===
 +
 
 +
$$ C_1 u - C_2 v = C_3 (u^2+v^2-R^2) $$
 +
for \(C_3\neq0\):
 +
$$ \frac{C_1}{C_3}u-\frac{C_1}{C_3}v=u^2+v^2-R^2 $$
 +
This looks like a circle, and indeed completing the square yields:
 +
$$ (u-\frac{C_1}{2C_3})^2 + (v+\frac{C_2}{2C_3})^2=\frac{C_1^2+C_2^2}{4C_3^2}+R^2 $$
 +
The geodesics in this case are circles centered at \( (\frac{C_1}{2C_3}, -\frac{C_2}{2C_3}) \) with radius \( \sqrt{\frac{C_1^2+C_2^2}{4C_3^2}+R^2} \). The radius is always greater than zero because  \(R>0\)
 +
 
 +
For \(C_3=0\):
 +
The geodesics are lines in \(\mathbb{R}^2\) parametrized by \( (C_2,C_1)t,  t\in\mathbb{R} \).
 +
 
 +
If  \(C_1 = C_3=0\), the equations do not give a unique solution  for \(u\) but \(v = 0\)
 +
 
 +
If  \(C_2 = C_3=0\), the equations do not give a unique solution for \(v\) but \(u = 0\)
 +
 
 +
If  \(C_1= C_2= C_3=0\):
 +
 
 +
$$C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v}) \Rightarrow v\dot{u} = u\dot{v}$$
 +
 
 +
$$ 0 = C_1  =\frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})$$
 +
 
 +
$$ 0 = 2uv\dot{u} +(R^2-u^2+v^2)\dot{v} = 2u^2\dot{v} +(R^2-u^2+v^2)\dot{v} = (R^2+u^2+v^2)\dot{v} \Rightarrow \dot{v} = 0$$
 +
 
 +
$$ C_2 = 0 \Rightarrow \dot{u} = 0$$
 +
 
 +
So in this case we have a geodesic that is constant at a single point in space.

Latest revision as of 06:10, 29 July 2015

Task

Let \(R > 0\). On \(\mathbb{R}^2\), consider the metric $$ g(u,v) = \frac{4R^4}{(R^2+u^2+v^2)^2} \begin{pmatrix} 1 & 0 \\ 0 & 1 \end{pmatrix},\ (u,v) \in \mathbb{R}^2$$

a) Write down the Hamiltonian equations describing the geodesic curves.

b) Let $$ C_1 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})$$ $$ C_2 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{v} +(R^2+u^2-v^2)\dot{u})$$ $$ C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v})$$ Show that \(C_1, C_2, C_3\) are conserved and that

$$ -C_1 u + C_2 v +C_3(u^2+v^2-R^2) = 0 \forall (u,v)\in \mathbb{R}^2$$ c) Show that the energy h satisfies \(h \geq 0\) and discuss the case \(h = 0\).

d) Describe the geodesics when \(h > 0\) depending on the values of \(C_1, C_2, C_3\).

Solution

Solution (a)

$$ g_{ij}(u,v) = \frac{4R^4}{(R^2+u^2+v^2)^2} \delta_{ij}$$

because \( \big(g^{ij}(u,v)\big) = \big(g_{ij}(u,v)\big)^{-1}\) $$ \Rightarrow g^{ij}(u,v) = \frac{(R^2+u^2+v^2)^2}{4R^4} \delta^{ij}$$

let \( x = (u,v)\) and \( \xi = (\xi_1, \xi_2)\), the hamiltonian is defined as $$h(x, \xi) = \frac{1}{2} g^{ij}(x)\xi_i \xi_j $$

$$ \Rightarrow h(x, \xi) = \frac{(R^2+u^2+v^2)^2}{8R^4} (\xi_1^2 + \xi_2^2)$$

now we can compute the Hamiltonian equations:

$$ \dot{u} = \frac{\partial h}{\partial \xi_1} = \frac{(R^2+u^2+v^2)^2}{4R^4} \xi_1$$ $$ \dot{v} = \frac{\partial h}{\partial \xi_2} = \frac{(R^2+u^2+v^2)^2}{4R^4} \xi_2$$

$$ \dot{\xi_1} = -\frac{\partial h}{\partial u} = -\frac{(R^2+u^2+v^2)}{2R^4} (\xi_1^2 + \xi_2^2) u$$

$$ \dot{\xi_2} = -\frac{\partial h}{\partial v} = -\frac{(R^2+u^2+v^2)}{2R^4} (\xi_1^2 + \xi_2^2) v$$

we can also observe that:

$$ \dot{\xi_2}u = \dot{\xi_1}v$$ and $$ \dot{u}\xi_2 = \dot{v}\xi_1$$


Solution (b)

$$ C_1 = \frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{u} +(R^2-u^2+v^2)\dot{v}) $$ using (a) $$ = 2uv\xi_1 +(R^2 -u^2 +v^2)\xi_2 = 2uv\xi_1 +R^2\xi_2 -u^2\xi_2 +v^2\xi_2$$

if \(C_1\) is conserved, its time derivative should be zero

$$\frac{dC_1}{dt} = 2\dot{u}v\xi_1 + 2u\dot{v}\xi_1 + 2uv\dot{\xi_1}+R^2\dot{\xi_2} - 2u\dot{u}\xi_2 - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$ using \(\dot{v}\xi_1= \dot{u}\xi_2\) on the second term, the second and fifth term cancel: $$ = 2\dot{u}v\xi_1 +2uv\dot{\xi_1}+R^2\dot{\xi_2} - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$ using \(v\dot{\xi_1} = u\dot{\xi_2}\) on the second term: $$ = 2\dot{u}v\xi_1 +2u^2\dot{\xi_2}+R^2\dot{\xi_2} - u^2\dot{\xi_2}+ 2v\dot{v}\xi_2 + v^2\dot{\xi_2}$$ $$ = \dot{\xi_2}(R^2 + u^2+ v^2) + 2\dot{u}v\xi_1 + 2v\dot{v}\xi_2 $$ using the Hamiltonian equations: $$ = -\frac{(R^2+u^2+v^2)^2}{2R^4}(\xi_1^2 + \xi_2^2) v + 2v\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_1^2 + 2v\frac{(R^2+u^2+v^2)^2}{4R^4} \xi_2^2 =0$$

As \(C_1, C_2\) and the hamilton equation are symmetric under permutation of \(u\) and \(v\), \(C_2\) is also conserved.

and for \( C_3 \)

$$ C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v}) = (v\xi_1 - u\xi_2)$$ $$ \frac{dC_3}{dt} = \dot{v}\xi_1 + v\dot{\xi_1} - \dot{u}\xi_2 - u\dot{\xi_2} = \dot{u}\xi_2 + v\dot{\xi_1} - \dot{u}\xi_2 - v\dot{\xi_1} = 0$$

Last we want to prove the identity:

$$ -C_1 u + C_2 v +C_3(u^2+v^2-R^2) = \frac{4R^4}{(R^2+u^2+v^2)^2}\Big(-(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})u $$ $$ + (2uv\dot{v} +(R^2+u^2-v^2)\dot{u})v + (v\dot{u} - u\dot{v})(u^2+v^2-R^2)\Big)$$ $$ = \frac{4R^4}{(R^2+u^2+v^2)^2}\Big( -2uv\dot{u}u -R^2\dot{v}u+u^2\dot{v}u-v^2\dot{v}u + 2uv\dot{v}v +R^2\dot{u}v+u^2\dot{u}v-v^2\dot{u}v$$ $$+ v\dot{u}u^2 + v\dot{u}v^2 - v\dot{u}R^2 - u\dot{v}u^2 - u\dot{v}v^2 + u\dot{v}R^2\Big) = 0$$ as all the terms in the second bracket cancel.

Solution (c)

$$ h(x, \xi) = \frac{(R^2+u^2+v^2)^2}{8R^4} (\xi_1^2 + \xi_2^2)$$ \(h \geq 0\) simply because all the terms are squared. For h = 0 the last bracket has to be zero because \(R > 0\) \( \Rightarrow \xi_1^2 = - \xi_2^2 \Rightarrow \xi_1 = \xi_2 = 0 \) from the Hamiltonian equation follows that u = const. , v = const. which corresponds to the case of a constant geodesic.


Solution (d)

$$ C_1 u - C_2 v = C_3 (u^2+v^2-R^2) $$ for \(C_3\neq0\): $$ \frac{C_1}{C_3}u-\frac{C_1}{C_3}v=u^2+v^2-R^2 $$ This looks like a circle, and indeed completing the square yields: $$ (u-\frac{C_1}{2C_3})^2 + (v+\frac{C_2}{2C_3})^2=\frac{C_1^2+C_2^2}{4C_3^2}+R^2 $$ The geodesics in this case are circles centered at \( (\frac{C_1}{2C_3}, -\frac{C_2}{2C_3}) \) with radius \( \sqrt{\frac{C_1^2+C_2^2}{4C_3^2}+R^2} \). The radius is always greater than zero because \(R>0\)

For \(C_3=0\): The geodesics are lines in \(\mathbb{R}^2\) parametrized by \( (C_2,C_1)t, t\in\mathbb{R} \).

If \(C_1 = C_3=0\), the equations do not give a unique solution for \(u\) but \(v = 0\)

If \(C_2 = C_3=0\), the equations do not give a unique solution for \(v\) but \(u = 0\)

If \(C_1= C_2= C_3=0\):

$$C_3 = \frac{4R^4}{(R^2+u^2+v^2)^2}(v\dot{u} - u\dot{v}) \Rightarrow v\dot{u} = u\dot{v}$$

$$ 0 = C_1 =\frac{4R^4}{(R^2+u^2+v^2)^2}(2uv\dot{u} +(R^2-u^2+v^2)\dot{v})$$

$$ 0 = 2uv\dot{u} +(R^2-u^2+v^2)\dot{v} = 2u^2\dot{v} +(R^2-u^2+v^2)\dot{v} = (R^2+u^2+v^2)\dot{v} \Rightarrow \dot{v} = 0$$

$$ C_2 = 0 \Rightarrow \dot{u} = 0$$

So in this case we have a geodesic that is constant at a single point in space.